Solutions — Parabolas 7B Worksheet
Year 10 Mathematics Core · Cambridge Ch 7, §7B · Mr Wong
ANSWER KEY
Part A — Reading the turning point
Q1. Turning points:
a $(0,3)$
b $(0,-4)$
c $(2,0)$
d $(-5,0)$
Q2. Turning point · max/min · axis · $y$-intercept:
a $y=(x-1)^2+2$: $(1,2)$ · min · $x=1$ · $(0,3)$
b $y=(x+4)^2-1$: $(-4,-1)$ · min · $x=-4$ · $(0,15)$
c $y=-(x-2)^2+5$: $(2,5)$ · max · $x=2$ · $(0,1)$
d $y=-(x+3)^2$: $(-3,0)$ · max · $x=-3$ · $(0,-9)$
Q3. Translations of $y=x^2$:
a up 3
b right 3
c left 1
d down 6
Part B — Reading a sketch
Q4. The graphed parabola $y=(x-2)^2-3$:
a minimum
b $(2,-3)$
c $x=2$
d $(0,1)$
Part C — Sketching from turning point form
Q5. Key features to mark on each sketch (curve should pass through the turning point and the $y$-intercept):
a $y=(x-1)^2-4$: minimum $(1,-4)$ · axis $x=1$ · $y$-intercept $(0,-3)$ · crosses $x$-axis at $-1$ and $3$.
b $y=-(x+1)^2+4$: maximum $(-1,4)$ · axis $x=-1$ · $y$-intercept $(0,3)$ · crosses $x$-axis at $-3$ and $1$.
Part D — Finding the rule
Q6. Substitute the given point into $y=ax^2+k$ (the turning-point height is $k$):
a $y=3x^2+2$ $5=a(1)^2+2 \Rightarrow a=3$.
b $y=2x^2-3$ $5=a(2)^2-3 \Rightarrow 4a=8 \Rightarrow a=2$.
Challenge
Q7. Maximum at $(2,5)$ through $(0,1)$:
a $h=2$, $k=5$ turning point form $y=a(x-2)^2+5$.
b $a=-1$ $1=a(0-2)^2+5 \Rightarrow 4a=-4 \Rightarrow a=-1$ (negative, so it is a maximum — consistent).
c $y=-(x-2)^2+5$